MathLabs

Problem 4

Show that the solution set of the inequality ∑k=170kx−k≥54\sum_{k=1}^{70}\frac{k}{x-k}\ge\frac{5}{4} is a union of disjoint intervals whose total length is 19881988.
Step 4 of 4: Apply Vieta's formula
Previous stepNext step
∑n=170rn=95(1+2+⋯+70)=9⋅7⋅71,L=28⋅71=1988\sum_{n=1}^{70}r_n=\frac{9}{5}(1+2+\cdots+70)=9\cdot7\cdot71,\qquad L=28\cdot71=1988
Detailed analysis

Let P(x)=∏k=170(x−k)P(x)=\prod_{k=1}^{70}(x-k). In 4∑k=170kP(x)/(x−k)−5P(x)=04\sum_{k=1}^{70}kP(x)/(x-k)-5P(x)=0, the leading coefficient is −5-5 and the coefficient of x69x^{69} is 9(1+2+⋯+70)9(1+2+\cdots+70). Thus Vieta gives ∑n=170rn=95(1+2+⋯+70)=9⋅7⋅71\sum_{n=1}^{70}r_n=\frac{9}{5}(1+2+\cdots+70)=9\cdot7\cdot71. Substitution into LL yields L=28⋅71=1988L=28\cdot71=1988.